$u$-substitute in integration

Question

So the $u$-substitute theorem states that $$\int f(g(x))g’(x)~dx = \int f(u)~du$$ But when we use the $u$-substitution, we say $u=g(x)$ right? So my question is why when we use this we can change $u$ and then take derivative. For example, if we do $u=3x^2-1$ than we can write $x^2=(u+1)/3$ And only then take derivatives from both sides? Plus what operation can we do before taking derivatives from both sides? How does it stands with the theorem?

Answer 1

the first part is right indeed let $u = g(x)$ so $$\frac{du}{dx} = g'(x)\implies du = g'(x)dx$$

And we can easily write $$\int f(g(x))g'(x)dx= \int f(u)du$$

You can check also on wikipedia I think it's explained very well - at least for me :) -

The second part is trying to reverse the process. Taking your example let $u=g(x)=3x^2-1$ so we have $$du = 6xdx\\ \frac{du}{6x} = dx $$ And what is $x$ in term of $u$? The inversion above so $$x= \pm\sqrt{\frac{u+1}{3}}$$ Here is where we need to be a little careful because the function we substitute has to be continuous. For the sake of simplicity I'll go with $+$ and $$dx= \frac{1}{6}\cdot \bigg(\frac{u+1}{3}\bigg)^{-1/2}du$$

and is exactly what we write above in terms of $u$. Hope this helps!